“玲珑杯”ACM比赛 Round #5 B -- Private Value【STL-map】_there are two octal integers numbers a and b.you n-优快云博客

本文链接：https://blog.youkuaiyun.com/hurmishine/article/details/53360157

本文介绍了一种求解私有值问题的算法，即在一个整数数组中找到出现频率最低的元素，并按升序输出这些元素。使用C++ map进行高效计数，通过迭代找出最小频次并输出所有符合条件的元素。

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B -- Private Value

Time Limit：1s Memory Limit：128MByte

Submissions：526Solved：174

DESCRIPTION

 The "private value" is the value that occurs, but most infrequent.
 An array of integers is given, and you need print the private values of the array.
 Please notice that, there might be several private values, and you need print them in increasing order.

INPUT

   The first line is an integer T(1 <= T <= 4), indicating the number of test cases.For each case, the first line contains an integer n (n <= 100000).The second line contains n integers a_i (0 <= a_i <= 10^9), 
  

OUTPUT

   For each case, print all private values in increasing order in a line.
  

SAMPLE INPUT

    2
   
    7
   
    1 1 2 2 3 3 3
   
    5
   
    5 3 2 4 1

SAMPLE OUTPUT

1 2

    1 2 3 4 5
   

SOLUTION

玲珑杯”ACM比赛 Round #5

原题链接：http://www.ifrog.cc/acm/problem/1057

题意：输出一个序列中出现次数最少的元素，数组可能会超时，类似于桶排序的方法估计也会超时，我用map就不用考虑那么多了。

AC代码：

/**
* 行有余力，则来刷题！
  * 博客链接:http://blog.youkuaiyun.com/hurmishine
  *
*/
#include <iostream>
#include <map>
using namespace std;
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        map<int,int>mp;
        mp.clear();
        int x;

        for(int i=0; i<n; i++)
        {
            cin>>x;
            mp[x]++;
        }
        map<int,int>::iterator  iter;
        int minn=1e9;
        for(iter=mp.begin(); iter!=mp.end(); iter++)
        {
            if(iter->second<minn)
                minn=iter->second;
        }
        bool first=true;
        for(iter=mp.begin(); iter!=mp.end(); iter++)
        {
            if(iter->second==minn)
            {
                if(!first)
                    cout<<" "<<iter->first;
                else
                {
                    first=false;
                    cout<<iter->first;
                }
            }
        }
        cout<<endl;
    }
    return 0;
}